Consider the titration of 100.0 mL of 0.200 M acetic acid () by 0.100 M . Calculate the pH of the resulting solution after the following volumes of have been added. 0.0 mL pH = 50.0 mL pH = 100.0 mL pH = 120.0 mL pH = 200.0 mL pH = 260.0 mL pH =

a) 2.72

b) 4.26

c) 4.74

d) 8.65

e) 8.79

f) 12.22

Explanation:

Considered the titration of 100.0 mL of 0.200 M acetic acid (Ka=1.8 x 10^-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

Step 1: Data given

Volume of 0.200 M acetic acid = 100.0 mL = 0.100 L

Concentration of KOH = 0.100M

Ka of acetic acid = 1.8 * 10^-5

pKa = 4.74

Step 2: The balanced equation

KOH + CH3COOH → CH3COOK + H2O

a) When 0.0 mL is added

pH = - log*(√[HA]*Ka)

pH = -log (√(0.200 * 1.8 x 10^-5)

pH = 2.72

b) When 50.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.0500 L * 0.100 M=0.005  moles

moles acetic acid in excess = 0.0200 - 0.005=0.015  moles

moles acetate = 0.005 moles

pH = pKa + log ([acetate]/[acetic acid])

pH = 4.74 + log 0.005/0.015= 4.26

c) When 100.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.100 L * 0.100 M=0.010 moles

moles acetic acid in excess = 0.0200 - 0.010=0.010 moles

moles acetate = 0.010 moles

CH3COO- + H2O <=> CH3COOH + OH-

pH = 4.74 + log 0.0100/0.0100= 4.74

d) When 120.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.120 L * 0.100 M=0.012 moles

moles acetic acid in excess = 0.0200 - 0.012=0.008  moles

moles acetate = 0.008 moles

total volume = 0.220 L

concentration acetate = 0.008/0.220=0.0364 M

CH3COO- + H2O <=> CH3COOH + OH-

Kb = 5.6 * 10^-10 =x^2/ 0.0364-x

x = [OH-]= 4.51 * 10^-6  M

pOH = 5.35

pH = 8.65

e)  When 200.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.200 L * 0.100 M=0.0200 moles

moles acetic acid in excess = 0.0200 - 0.0200 = 0  moles

moles acetate = 0.020 moles

total volume = 0.300 L

concentration acetate = 0.020/0.300 =0.0667 M

CH3COO- + H2O <=> CH3COOH + OH-

Kb = 5.6 * 10^-10 =x^2/ 0.0667-x

x = [OH-]=6.11 * 10^-6  M

pOH = 5.21

pH = 8.79

f)  When 260.0 mL KOH is added

moles acetic acid = 0.100 L * 0.200 M = 0.0200  moles

moles OH- added = 0.260 L * 0.100 M=0.0260 moles

moles OH- in excess = 0.0260 - 0.0200 = 0.0060 moles

total volume = 0.360 L

[OH-] = 0.0060 moles / 0.360 L

[OH-]=  0.0167

pOH = -log [0.0167]

pOH = 1.78

pH = 14- 1.78 = 12.22

to make a supersaturated solution, make a saturated solution of sugar by adding 360 grams of sugar to 100 ml of water at 80 degrees celsius. when the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water should only dissolve 210 grams of sugar.

answer: h+ its less than oh

explanation: